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We have 100 bags which contains finite number of colored balls. None of them are empty.

We gonna put them into "k" boxes.

What is the minimum value of "k" which no matter how the contents of bags are arranged the bags can be distributed into k boxes so that for each box at least one of the following two conditions is held:

• all bags of a box contain a ball of the same color

• each bag of a box contains a ball colored differently from all balls of all other bags of this box.

I honestly don't have any idea to approach this one. For only first condition k must be 100 or higher.

For only second the highest value of k is 100.

But probably this is not the way I should've started approaching this so anyways.

Thanks for your help.

Cashews
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  • how many colors are there what are they distributed like ? –  Aug 20 '17 at 23:08
  • As questions says it is changeable. We can put any number or color of balls to 100 bags. And k boxes should still be enough. – Cashews Aug 20 '17 at 23:10

1 Answers1

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$50$ boxes with $2$ bags per box works - either of the conditions will be satisfied (trivial to show).

However, if there are $49$ boxes, then by the pigeonhole principle, at least one box must contain $3$ bags. It is then possible to construct an example where both conditions are failed in that box. Eg. the bags in the box are $\{A_1,B_1\},\{A_2,C_1\},\{B_2,C_2\}$ (where $A_x,B_x,C_x$ are balls with colors $A,B,C$).

Hence $50$ is the minimum value.

I think your initial approach is the right one. You notice that the conditions imply that $k<100$. From there, you might think of various values of $k$ to consider (the answer came to me when I decided to consider $100\div2=50$ and noting the PHP might be used in some way).

Shuri2060
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